Php 如何模拟';查找';symfony3中的方法
我试图模拟Php 如何模拟';查找';symfony3中的方法,php,symfony,testing,doctrine-orm,mocking,Php,Symfony,Testing,Doctrine Orm,Mocking,我试图模拟EntityRepository的find方法,这样测试就不会在数据库中查找数据,但似乎不起作用。下面是测试类的设置方法 public function setUp() { parent::setUp(); $this->client = static::createClient(); $this->peopleManager = $this->getMockBuilder(PeopleManager::class) ->
EntityRepository
的find
方法,这样测试就不会在数据库中查找数据,但似乎不起作用。下面是测试类的设置
方法
public function setUp()
{
parent::setUp();
$this->client = static::createClient();
$this->peopleManager = $this->getMockBuilder(PeopleManager::class)
->setMethods(['createPerson','peopleUpdate', 'peopleDelete', 'peopleRead'])
->disableOriginalConstructor()
->getMock();
$this->repository = $this->getMockBuilder(EntityRepository::class)
->disableOriginalConstructor()
->getMock();
$this->em = $this->getMockBuilder(EntityManager::class)
->disableOriginalConstructor()
->getMock();
}
这就是我们调用find函数的方法
public function updatePersonAction($id, Request $request)
{
$repository = $this->getDoctrine()->getRepository('GeneralBundle:People');
$person= $repository->find($id);
if($person)
{
$data = $request->request->get('array');
$createdPeople = array();
$UpdatedPerson = "";
foreach($data as $content)
{
$prueba = $this->get('people.manager');
$UpdatedPerson = $prueba->peopleUpdate(
$person,
$content['name'],
$content['surname'],
$content['secondSurname'],
$content['nationality'],
$content['birthday'],
$content['identityCard'],
$content['identityCardType']
);
array_push($createdPeople, $person);
}
$serializedEntity = $this->get('serializer')->serialize($UpdatedPerson, 'json');
return new Response($serializedEntity);
} else {
$serializedEntity = $this->get('serializer')->serialize('Doesn\'t exists any person with this id', 'json');
return new Response($serializedEntity);
}
}
调试器显示peoplemanager类是模拟的,但它不会模拟实体管理器和存储库
谢谢假设您要测试的类如下所示:
// src/AppBundle/Salary/SalaryCalculator.php
namespace AppBundle\Salary;
use Doctrine\Common\Persistence\ObjectManager;
class SalaryCalculator
{
private $entityManager;
public function __construct(ObjectManager $entityManager)
{
$this->entityManager = $entityManager;
}
public function calculateTotalSalary($id)
{
$employeeRepository = $this->entityManager
->getRepository('AppBundle:Employee');
$employee = $employeeRepository->find($id);
return $employee->getSalary() + $employee->getBonus();
}
}
由于ObjectManager通过构造函数注入到类中,因此很容易在测试中通过模拟对象:
// tests/AppBundle/Salary/SalaryCalculatorTest.php
namespace Tests\AppBundle\Salary;
use AppBundle\Entity\Employee;
use AppBundle\Salary\SalaryCalculator;
use Doctrine\ORM\EntityRepository;
use Doctrine\Common\Persistence\ObjectManager;
use PHPUnit\Framework\TestCase;
class SalaryCalculatorTest extends TestCase
{
public function testCalculateTotalSalary()
{
// First, mock the object to be used in the test
$employee = $this->createMock(Employee::class);
$employee->expects($this->once())
->method('getSalary')
->will($this->returnValue(1000));
$employee->expects($this->once())
->method('getBonus')
->will($this->returnValue(1100));
// Now, mock the repository so it returns the mock of the employee
$employeeRepository = $this
->getMockBuilder(EntityRepository::class)
->disableOriginalConstructor()
->getMock();
$employeeRepository->expects($this->once())
->method('find')
->will($this->returnValue($employee));
// Last, mock the EntityManager to return the mock of the repository
$entityManager = $this
->getMockBuilder(ObjectManager::class)
->disableOriginalConstructor()
->getMock();
$entityManager->expects($this->once())
->method('getRepository')
->will($this->returnValue($employeeRepository));
$salaryCalculator = new SalaryCalculator($entityManager);
$this->assertEquals(2100, $salaryCalculator->calculateTotalSalary(1));
}
}
在本例中,您从内到外构建模拟,首先创建由存储库返回的employee,该employee本身由EntityManager返回。这样,测试中就不会涉及到真正的类
资料来源:
对于mocky来说,这是非常微不足道的,尝试一下类似于
$repo->shouldReceive('find')->once()->andReturn([1,2,3])代码>谢谢,这是非常有用的!!